Question Video: Finding the Indefinite Integral of a Function Containing Exponential Functions by Distributing the Division Mathematics

Video Transcript

Determine the indefinite integral of eight times 𝑒 to the power of three 𝑥 minus 𝑒 to the power of two 𝑥 plus nine divided by seven 𝑒 to the power of 𝑥 with respect to 𝑥.

In this question, we’re asked to evaluate the indefinite integral of an exponential expression. And at first, it might look very difficult to evaluate this integral. For example, we might be considering a 𝑢-substitution. However, we should always check if we can simplify our integrand. And we can do this by dividing every term in the numerator by the denominator, where we remember for each term in the numerator, we just need to subtract the exponents of 𝑒. This gives us the indefinite integral of eight over seven times 𝑒 to the power of two 𝑥 minus one-seventh 𝑒 to the power of 𝑥 plus nine over seven times 𝑒 to the power of negative 𝑥 with respect to 𝑥.

Now, we can see we know how to integrate each term in the integrand separately. So, we’re going to split this into three separate integrals. And we can do this because we recall the integral of the sum of functions is equal to the sum of their individual integrals. This gives us the following. We can then simplify each term separately. We’ll take the constant factor outside of each integral. This then gives us eight-sevenths times the indefinite integral of 𝑒 to the power of two 𝑥 with respect to 𝑥 minus one-seventh the indefinite integral of 𝑒 to the power of 𝑥 with respect to 𝑥 plus nine-sevenths the integral of 𝑒 to the power of negative 𝑥 with respect to 𝑥.

We’re now ready to evaluate each of these indefinite integral separately. We can do this by recalling for any real constant 𝑎 not equal to zero, the integral of 𝑒 to the power of 𝑎𝑥 with respect to 𝑥 is one over 𝑎 times 𝑒 to the power of 𝑎𝑥 plus the constant of integration 𝐶. We’ll use this to evaluate each integral separately. Let’s start with the first integral, the integral of 𝑒 to the power of two 𝑥 with respect to 𝑥. Our value of 𝑎 is two. So, we substitute 𝑎 is two into our integral rule. We get one-half times 𝑒 to the power of two 𝑥. And we need to multiply this by eight over seven. And it’s worth noting since we’re evaluating the sum of multiple indefinite integrals, we’ll get a constant of integration for each of these integrals. We can combine all of these constants into one constant at the end of our expression. So, we don’t need to worry about adding 𝐶 until the end.

Let’s now evaluate our second integral. We know 𝑒 to the power of 𝑥 is the same as 𝑒 to the power of one times 𝑥. So, our value of 𝑎 will be one. Of course, this is somewhat not necessary since we know the integral of 𝑒 to the power of 𝑥 with respect to 𝑥 is just 𝑒 to the power of 𝑥 with respect to 𝑥. However, if we did substitute 𝑎 is equal to one, we would get one over one times 𝑒 to the power of one 𝑥, which is just 𝑒 to the power of 𝑥. So, evaluating our second integral and multiplying by negative one-seventh, we get negative one-seventh 𝑒 to the power of 𝑥.

Finally, let’s evaluate the third integral. Our value of 𝑎 is negative one. Substituting 𝑎 is negative one into our integral result gives us one over negative one times 𝑒 to the power of negative 𝑥. We need to multiply this by nine over seven. And remember, we also need to add a constant of integration at the end of this expression. We can then simplify this slightly. One over negative one is just equal to negative one. So, instead of adding this term, we can subtract this term. And we can also notice in the first term, we have a shared factor of two in the numerator and denominator. We can cancel this to get a factor of four over seven.

This then gives us our final answer. The indefinite integral of 𝑒 to the power of three 𝑥 minus 𝑒 to the power two 𝑥 plus nine over seven 𝑒 to the 𝑥 with respect to 𝑥 is equal to four over seven times 𝑒 to the power of two 𝑥 minus 𝑒 to the power of 𝑥 over seven minus nine-sevenths 𝑒 to the power of negative 𝑥 plus the constant of integration 𝐶.